Mathematics - Area Under Curves Question with Solution | TestHub

The required area is given by the definite integral of the curve $y=|x-2|$ from $x=1$ to $x=3$.

$\text{Area}={\int }_1^3|x-2|dx$

Since $|x-2|=2-x$ for $1\le x\le 2$ and $|x-2|=x-2$ for $2\le x\le 3$, we split the integral at $x=2$:

$\text{Area}={\int }_1^2(2-x)dx+{\int }_2^3(x-2)dx$

Evaluating the integrals:

$\text{Area}={\left[2x-\frac{x^2}{2}\right]}_1^2+{\left[\frac{x^2}{2}-2x\right]}_2^3$

$\text{Area}=\left((4-2)-\left(2-\frac{1}{2}\right)\right)+\left(\left(\frac{9}{2}-6\right)-(2-4)\right)$

$\text{Area}=\left(2-\frac{3}{2}\right)+\left(-\frac{3}{2}+2\right)$

$\text{Area}=\frac{1}{2}+\frac{1}{2}=1\text{ sq. unit}$

Answer: 1 sq.units